思路是dfs+剪枝 sum是前缀和，ans是答案，le是还剩下多少重量 两个剪枝： ①砍瓜次数大于等于已知最优剪枝 ②剩余全部可选瓜重量之和（sum记录）超过所需重量剪枝

11号测试点WA，应输出4，程序输出6。 求指教

#include<iostream>
#include<queue>
#include<algorithm>
#include<stdlib.h>
#include<time.h> 
#define int long long
using namespace std;
int n, m, sum[100];
int a[100];
int ans = 31;
void dfs(int now, int le, int cnt) {
	if (cnt >= ans || le > (sum[n] - sum[now - 1]))return;
	if (le == 0) {
		ans = cnt;
		return;
	}
	for (int i = now; i <= n; i++) {
		while (a[i] == a[i - 1])i++;
		if (le >= a[i] / 2) {
			dfs(i + 1, le - a[i] / 2, cnt + 1);
		}
		if (le >= a[i]) {
			dfs(i + 1, le - a[i], cnt);
		}
	}
}
signed main() {
	cin >> n >> m;
	m = m * 2;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		a[i] *= 2;
	}
	sort(a + 1, a + n + 1);
	reverse(a + 1, a + n + 1);
	for (int i = 1; i <= n; i++)sum[i] = sum[i - 1] + a[i];
	dfs(1, m, 0);
	if (ans <= 30) {
		cout << ans << endl;
	}
	else  cout << -1 << endl;
}